How do you find the vertex and intercepts for #f(x)=4x^2-32x+63#?

1 Answer
Dec 10, 2015

I have given the answer to and shown the method to obtain
#color(blue)(x_("vertex")= +4)#
I have given the method to find the rest.

Explanation:

Tony B

#color(blue)("To find " x_("vertex"))#

Given: #y=4x^2-32x+63 ........(1)#

Write as: #4(x^2-32/4 x)+63#

Consider the #-32/4 " from " -32/4x#

#x_("vertex")=(-1/2)(-32/4) = +32/8 =4...(2)#
This compares to the graph
.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Method from this point:")#

#color(blue)("To find " y_("vertex"))#

Substitute (2) into (1) to solve for #y_("vertex")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To find " y_("intercept"))#

#y_("intercept") = 63# this is the constant at the end of equation (1)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("To find " x_("intercept"))#

Substitute #y=0# in equation (1) and solve for #x#

If you are not sure about factoring use the formula

Standard form: #y=ax^2+bx+c#

where #x=(-b+-sqrt(b^2-4ac))/(2a)#

You can see from the graph that your answers should be close to #3 1/2 " and " 4 1/2 color(red)(" "underline("These are estimates"))#